You can
put this solution on YOUR website!If you have drawn a diagram of this situation, then it should look like a right triangle in which the string is the hypotenuse, the height of the kite is the height of the triangle, and the base of the triangle is the horizontal distance from the person to the perpendicular line from the kite to the ground, (the height). To solve this, you will need to use the Pythagorean theorem: c^2 = a^2 + b^2
Where: c is the hypotenuse (string length of 50 ft), a is the perpendicular height of the kite above the ground, and b (let's call this x)) is the horizontal distance from the person the the perpendicular line from the kite to the ground. The height is x + 10 ft. Now you can set up the equation:
50^2 = (x+10)^2 + x^2
2500 = (x^2 + 20x + 100) + x^2 Simplify.
2500 = 2x^2 + 20x + 100 Divide through by 2 to facilitate calculations.
1250 = x^2 + 10x + 50 Subtract 1250 from both sides.
x^2 + 10x + 50 - 1250 = 0 Simplify.
x^2 + 10x - 1200 = 0 Solve this quadratic by factoring.
(x + 40)(x - 30) = 0 Apply the zero products principle.
x + 40 = 0, x = -40 Discard this solution as a negative height is not meaningful.
x - 30 = 0, x = 30 ft. This solution is acceptable.
Now this is the horizontal distance from the person. The height is this plus 10 ft.
So the height of the kite is: 30 ft. + 10 ft. = 40 ft.
You can
put this solution on YOUR website!My hat is off to Earlsdon, who apparently was solving this problem at the same time as I was, and who submitted the solution before I did!!
I can't draw the triangle, but the kite string is 50 feet, which is the hypotenuse of the right triangle.
Let x = the horizontal distance
x + 10 = the height of the kite (that even rhymes!!)
The legs of the right triangle are x and x+10, with the hypotenuse = 50.
By Theorem of Pythagoras,
This is quadratic, so combine like terms and subtract 2500 from each side of the equation to set it equal to zero:
Divide both sides by 2:
Factor the trinomial:
x = -40 or x = 30
You can't have a negative side of a triangle, so reject the -40. The answer is x = 30 for the horizontal distance, and x+10 = 40 feet for the height.
R^2 at SCC